POJ1003-hangOver
How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.
最初は、いろいろなアルゴリズムを試してみたが、まだ難しいと感じていたが、直接暴力で解決すればいいことに気づいた.時間は0 MSだったが、私はやはり考えすぎた.
最初は、いろいろなアルゴリズムを試してみたが、まだ難しいと感じていたが、直接暴力で解決すればいいことに気づいた.時間は0 MSだったが、私はやはり考えすぎた.
/*
*hang over
*POJ 1003
*author:yuan tian
*whu eis
* 0.00 - 0.52 ,
*/
#include
using namespace std;
int main()
{
int cardIndex = 0;
double queryNum = 0.0;
double overHang = 0.0;
while (cin >> queryNum)
{
overHang = 0.0;
if (!queryNum)
break;
for (cardIndex = 1;; cardIndex++)
{
overHang += 1.0 / (double)(cardIndex + 1);
if (overHang > queryNum)
{
break;
}
}
cout << cardIndex << " card(s)" << endl;
}
}